Upper central series members are completely divisibility-closed in nilpotent group

Statement

Suppose $G$ is a nilpotent group. Then, the members of the upper central series of $G$ (with the possible exception of the zeroth member, the trivial subgroup) are completely divisibility-closed subgroups of $G$. Explicitly, this means that if $p$ is a prime number such that $G$ is $p$-divisible, then all the members of the upper central series of $G$ (possibly excepting the zeroth member, the trivial subgroup) are completely $p$-divisibility-closed subgroups of $G$.

In particular, the center and second center are both completely divisibility-closed subgroups, and hence also divisibility-closed subgroups.

Related facts

Similar facts

• Lower central series members are divisibility-closed in nilpotent group

Opposite facts

• Lower central series members need not be completely divisibility-closed in nilpotent group
• Center not is divisibility-closed

Facts similar to proof technique

• Epicentral series members are completely divisibility-closed in nilpotent group
• Annihilator of divisibility-closed subgroup under bihomomorphism is completely divisibility-closed
• Complete divisibility-closedness is strongly intersection-closed

Facts used

1. Annihilator of divisibility-closed subgroup under bihomomorphism is completely divisibility-closed
2. Complete divisibility-closedness is strongly intersection-closed
3. Complete divisibility-closedness is transitive

Proof

The proof of this is unusual in that the induction proceeds downward from the penultimate member (the largest). The idea is to shift the "take the $p^{th}$ root" operation between the elements in the iterated commutator map and note that the output is unaffected. This forces all $p^{th}$ roots of an element in the appropriate upper central series member to be in that upper central series member.